proof of continuous functions are Riemann integrable

Recall the definition of Riemann integral. To prove that $f$ is integrable we have to prove that $\lim_{\delta\to 0^{+}}S^{*}(\delta)-S_{*}(\delta)=0$ . Since $S^{*}(\delta)$ is decreasing and $S_{*}(\delta)$ is increasing it is enough to show that given $\epsilon>0$ there exists $\delta>0$ such that $S^{*}(\delta)-S_{*}(\delta)<\epsilon$ .

So let $\epsilon>0$ be fixed.

By Heine-Cantor Theorem $f$ is uniformly continuous i.e.

\exists\delta>0\ |x-y|<\delta\Rightarrow|f(x)-f(y)|<\frac{\epsilon}{b-a}.

Let now $P$ be any partition of $[a,b]$ in $C(\delta)$ i.e. a partition $\{x_{0}=a,x_{1},\ldots,x_{N}=b\}$ such that $x_{i+1}-x_{i}<\delta$ . In any small interval $[x_{i},x_{i+1}]$ the function $f$ (being continuous) has a maximum $M_{i}$ and minimum $m_{i}$ . Since $f$ is uniformly continuous and $x_{i+1}-x_{i}<\delta$ we have $M_{i}-m_{i}<\epsilon/(b-a)$ . So the difference between upper and lower Riemann sums is

\sum_{i}M_{i}(x_{i+1}-x_{i})-\sum_{i}m_{i}(x_{i+1}-x_{i})\leq\frac{\epsilon}{b% -a}\sum_{i}(x_{i+1}-x_{i})=\epsilon.

This being true for every partition $P$ in $C(\delta)$ we conclude that $S^{*}(\delta)-S_{*}(\delta)<\epsilon$ .

Title	proof of continuous functions are Riemann integrable
Canonical name	ProofOfContinuousFunctionsAreRiemannIntegrable
Date of creation	2013-03-22 13:45:34
Last modified on	2013-03-22 13:45:34
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	7
Author	paolini (1187)
Entry type	Proof
Classification	msc 26A42